简便方法计算：一、 (1又1998分之1乘8.12加5又75分之31除以3分之2乘8又1998分之1997)除以4又48

1又3分之2减[(3.62乘15分之2加6.38除以7又2分之1)乘1.82减1又3分之1乘0.82]
解原式=5/3-[(362/100×2/15+638/100×2/15)×182/100-4/3×82/100]
=5/3-[(362/100+638/100)×2/15×182/100-4/3×82/100]
=5/3-(10×2/15×182/100-4/3×82/100)
=5/3-(4/3×182/100-4/3×82/100)
=5/3-[4/3(182/100-82/100)]
=5/3-(4/3×1)
=5/3-4/3
=1/3
再问： 第一题呢
再答： (1又1998分之1乘8.12加5又75分之31除以3分之2乘8又1998分之1997)除以4又48分之11 =[(1999/1998）×812/100+（406/75)÷(2/3)×(1997+8×1998)/1998]÷203/48 =[(1999/1998)×203/25+203/25×(1997+8×1998)1998]÷203/48 =1/25×(1999/1998+1997/1998+8)×48 =48/25×(2+8) =48×10÷25 =19.2